KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4

Question 1.
The sides of a rectangle are in the ratio 2: 1. The perimeter is 30 cm. Calculate the measure of all the sides.
Answer:
The ratio of the sides is 2 : 1
Let the sides be 2x and x
Perimeter = 30
2x + x + 2x + x = 30
6x = 30
x = \(\frac{3}{6}\) = 5 cm
2x = 2 x 5 = 10cm
∴ The sides aer 10 cm, 5 cm, 10 cm, 5 cm.

Question 2.
In the adjacent rectangle ABCD ∠OCD = 30° . Calculate ∠BOC. What type of triangle is BOC.
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.4 1
Answer:
∠BCD = 90° [Angle of a rectangle]
∠OCD + ∠OCB = 90°
30 + ∠QCB = 90°
∠OCB = 90 – 30
∠OCB = 60°
∠OCB = ∠OBC = 60°
OC = OB [Diagonals of a rectangle bisect each other]
∴ ∠BOC = 180 – (60 + 60) = 180 – 120
∠BOC = 60°
In Δ BOC, ∠BOC = ∠OBC = ∠OCB = 60° & equiangular triangle
∴ BOC is an equiangular triangle.

Question 3.
All the rectangles are parallelograms but all the parallelograms are not rectangles. Justify this statement.
Answer:
A rectangle has all the properties of a parallelogram, therefore, all rectangles are parallelograms. No angle of a Parallelogram is a right angle but all angles of a rectangle are right angles.
∴ All parallelograms are not rectangular.

Question 4.
The side of a rectangular park are in the ratio 4 : 3. If the area is 1728 m find the cost of fencing it at the rate of Rs . 2.50/m
Answer:
The ratio of the sides = 4 : 3
Let the sides be 4x and 3x
Area = 1728
l × b = 1728
4x × 3x = 1728
12x2 = 1728
x2 = \(\frac{12}{1728}\)
x2 = 144
x = √144 = 12m
4x = 4 × 12 = 48m
3x = 3x × 12 = 36m
Perimeter = 4x + 3x + 4x + 3x
= 48 + 36 + 48 + 36
= 168m
The cost of fencing 1m is Rs 2.50
The cost of fencing 168m = 168 × 250 = Rs. 420.

Question 5.
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remaining portion is a yard of a trapezoidal shape whose parallel sides have length 15m and 25m. What fraction of the yard is occupied by the flower bed?
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.4 2
Answer:
In the fig AB = 25 m and FE = 15 m
∴ DF + EC = 25 – 15 = 10m
But DF = EC
∴DF = EC = 5m also AD = BC = 5m
Area of rectangle ∆BCD = AB x BC = 1 x b = 25 x 5 = 125 m2
Area of ∆ADF = \(\frac{1}{2}\) base × height
= \(\frac{1}{2}\) 5 × 5 = \(\frac{25}{2}\)m2
∆ADF ≅ ABCE [data]
Area of ∆BCE = \(\frac{25}{2}\) m2
Sum of Area of the two triangles
∴ \(\frac{1}{5}\) of the field is occupied by the flower bed.

Question 6.
In a rhombus ABCD  ∠C = 70°. Find the other angles of the rhombus.
Answer:
∠A = ∠C = 70° [Opposite angles]
∠A +∠B = 180° [Adjacent angles of a rhombus]
70 + ∠B = 180°
∠B = 180 – 70
∠B = 110°
∠B = ∠D = 110° [Opposite angles are equal].
Angles are 70°, 110°, 70° & 110°

Question 7.
In a rhombus PQRS, if PQ = 3x – 7 and QR = x + 3 find PS.
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.4 3
Answer:
PQ = QR [Sides of a rhombus]
3x – 7 = x + 3
3x – x = 3 + 7
2x = 10
\(\frac{10}{2}\)
x = 5
PQ = 3x – 7 = 3 × 5 – 7 = 15 – 7
PQ = 8 cm
∴ SP = 8

Question 8.
Rhombus is a parallelogram. Justify.
Answer:
In a rhombus opposite sides are equal and parallel to each other therefore it is a parallelogram.

Question 9.
In a given square ABCD, if the area of triangle ABD is 36cm2 Find
1. The area of the triangle BCD and
2. Area of the square ABCD
Answer:
The diagonal of a square divides the square into two congruent triangles.
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.4 4
∴ Area of ∆ABD = ∆BCD = 36cm²
Area of square ABCD = 36 + 36 = 72 cm².

Question 10.
The side of square ABCD is 5cm and another square PQRS has a perimeter equal to 40cm. Find the ratio of perimeter ABCD to the perimeter of PQRS. Find the ratio of the area of ABCD to the area of PQRS.
Answer:
Side of ABCD = 5 cm
Perimeter of ABCD = 5 × 4 = 20 cm
Perimeter of PQRS = 40 cm 4 × side = 40
side = \(\frac { 40 }{ 4 }\)
side = 10cm
Perimeter of ABCD : Perimeter PQRS
= 20 : 40 = 1 : 2
Area of ABCD:Area of PQRS
= 52 : 102 = 25 : 100 = 1 : 4

Question 11.
A square field has a side of 20m. Find the length of the wire required to fence it four times.
Answer:
Perimeter = 4a = 4 × 20 = 80m
Length of the wire required to fence 4 times. = 4 × 80 = 320 m.

Question 12.
List out the difference between square and rhombus
Answer:

Square Rhombus
i. All the angles are equal. i. Opposite angles are equal
ii. All the angles are right angles ii. No angle is a right angle
iii. Diagonals are equal iii. Diagonals are not equal.

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